# Acceleration as the logarithmic derivative of the redshift factor $U^\sigma \nabla_\sigma U^\mu = g^{\mu\nu}\nabla_\nu \ln V$

Let $$K^\mu = V(x)U^\mu$$ be vector proportional to the four-velocity $$U^\mu$$. (e.g. $$K^\mu$$ is a normalized time-like Killing vector for an observer at infinity).

Then, $$V(x) = \sqrt{-K_\nu K^\nu}$$ is called the "redshift factor".

Also, the four-acceleration $$a^\mu$$ is given by $$U^\sigma \nabla_\sigma U^\mu$$.

According to Sean Carroll, Spacetime and Geometry, p. 247, $$a_\mu = \nabla_\mu\ln V$$. Why?

Attempt:

\begin{align}\nabla_\mu\ln V &= \frac 1{2V^2}\nabla_\mu\left(-K_\nu K^\nu\right)\\ &=-\frac 1{2V^2}\left((\nabla_\mu K_\nu)K^\nu + K_\nu\nabla_\mu K^\nu\right)\\ &=-\frac 1{2V^2}\left((\nabla_\mu g_{\rho\nu}K^\rho)K^\nu + K_\nu\nabla_\mu K^\nu\right)\\ &=-\frac 1{2V^2}\left(K_\rho\nabla_\mu K^\rho + K_\nu\nabla_\mu K^\nu\right)\\ &=-\frac {K_\nu\nabla_\mu K^\nu}{V^2}\\ &= -\frac 1{V}U_\nu\nabla_\mu\left(VU^\nu\right)\\ &= -U_\nu\nabla_\mu U^\nu - \frac 1 VU_\nu U^\nu\nabla_\mu V\end{align}

At some point in your calculation, you got the following:

$$\nabla_\mu \ln{V} =-\frac{1}{V^2} K_\nu \nabla_\mu K^\nu$$

Using the metric compatibility, we can write

$$\nabla_\mu \ln{V} =-\frac{1}{V^2} K_\nu \nabla_\mu \left( g^{\nu \lambda} K_\lambda \right)=-\frac{1}{V^2} K_\nu g^{\nu \lambda} \nabla_\mu K_\lambda=-\frac{1}{V^2} K^\lambda \nabla_\mu K_\lambda.$$

Now, since $$K^\mu$$ is a Killing vector field, then it satisfies the Killing equation:

$$\nabla_\mu K_\lambda + \nabla_\lambda K_\mu =0.$$

Substituting in the equation above, we get

$$\nabla_\mu \ln{V}= \frac{1}{V^2} K^\lambda \nabla_\lambda K_\mu.$$

Using the relation $$K^\lambda= V(x) U^\lambda$$, we get

$$\nabla_\mu \ln{V}= \frac{1}{V^2} \left( VU^\lambda \right) \nabla_\lambda \left( V U_\mu \right).$$

Applying the Leibniz rule,

$$\nabla_\mu \ln{V}= \frac{1}{V} U^\lambda \left( (\nabla_\lambda V) U_\mu + (\nabla_\lambda U_\mu) V \right).$$

$$\therefore \nabla_\mu \ln{V}= U^\lambda U_\mu \frac{1}{V} \nabla_\lambda V + U^\lambda \nabla_\lambda U_\mu$$

Or, using $$a_\mu = U^\lambda \nabla_\lambda U_\mu$$ and $$\frac{1}{V} \nabla_\lambda V =\nabla_\lambda \ln{V}$$,

$$\nabla_\mu \ln{V}= U_\mu U^\lambda \nabla_\lambda \ln{V} + a_\mu \ \ \ \ \ \rightarrow (1)$$

Now, contract both side of equation (1) with $$U^\mu$$ and use the fact that $$a_\mu U^\mu=0$$ and $$U^\mu U_\mu=-1$$.

$$\therefore U^\mu \nabla_\mu \ln{V}= - U^\lambda \nabla_\lambda \ln{V}.$$

The only way this is satisfied is to have

$$U^\lambda \nabla_\lambda \ln{V}=0 \ \ \ \ \ \rightarrow (2)$$

Substitute from equation (1) into equation (2), you get:

$$\nabla_\mu \ln{V}=a_\mu,$$

which is what we wanted to prove.

• For those wondering, $U^\mu a_\mu = 0$ because $U^\mu \propto K^\mu$, so in one hand $U^\mu U^\nu = U^\nu U^\mu$, which implies that $U^\mu a_\mu = U^\mu U^\nu \nabla_\mu U_\nu$ after renaming indices, and in the other hand $\nabla_{(\mu}U_{\nu)} = 0$, which implies that $U^\mu a_\mu = -U^\mu U^\nu \nabla_\mu U_\nu$. This is only consistent iff $U^\mu a_\mu = 0$. Commented Dec 11, 2023 at 18:22